Problem: Factor completely. $3x^2+33x+90=$
Solution: First, let's factor out the greatest common factor. If the leading coefficient is negative, we'll also factor out $-1$. The result will be something like this, where the blanks are coefficients: $_{\llcorner\!\lrcorner}\!(_{\llcorner\!\lrcorner}\! x^2+\, _{\llcorner\!\lrcorner}\! x+\, _{\llcorner\!\lrcorner}\!)$ Then, we can try to factor the sum like this: $_{\llcorner\!\lrcorner}\!(_{\llcorner\!\lrcorner}\! x+\, _{\llcorner\!\lrcorner}\!) (_{\llcorner\!\lrcorner}\! x+\, _{\llcorner\!\lrcorner}\!)$ Factor out a the greatest common factor The greatest common factor is $3$. $\begin{aligned} &\phantom{=}3x^2+33x+90 \\\\ &=3(x^2+11x+30) \end{aligned}$ [How did we find the greatest common factor?] Factor $x^2+11x+30$ $\begin{aligned} &\phantom{=}3x^2+33x+90 \\\\ &=3(x^2+11x+30) \\\\ &=3(x+5)(x+6) \end{aligned}$ [I want to see this step in more detail.] Answer $\begin{aligned} &\phantom{=}3x^2+33x+90 \\\\ &=3(x+5)(x+6) \end{aligned}$